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4 years ago

Is muddy water a substance?

Physics

2 answers:

sergejj [24]4 years ago

7 0

Answer: No, I wouldn't imagine so.

Explanation:

Kazeer [188]4 years ago

7 0

Muddy water is not a substance

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The Nardo ring is a circular test track for cars. It has a circumference of 12.5 km. Cars travel around the track at a constant

KIM [24]

Answer: 50 km, 0, 27.78 m/s

Explanation:

Given

Circumference of the track is $12.5\ km$

Speed of car is $100\ km/h$

Car drives for $30\ \text{minute}\ or\ 0.5\ hr$

(a)Distance traveled is

$\Rightarrow D=100\times 0.5\\\Rightarrow D=50\ km$

(b)displacement of the car

It can be observed that 12.5 is a multiple of 50, that is, 50 km can be interpreted as 4 complete rounds of the track.

Therefore, the displacement of the car is zero.

(c)To convert kmph to m/s, multiply the entity by $\frac{5}{18}$

$\Rightarrow 100\times \dfrac{5}{18}\\\\\Rightarrow 27.78\ m/s$

5 0

3 years ago

The speed of all electromagnetic waves is 3.00 × 108 meters per second. What is the wavelength of an X-ray with a frequency of 1

Ierofanga [76]

The answer is......2.54x10^-10......

8 0

4 years ago

Read 2 more answers

As a car accelerates its tires increase their angular velocity from 114 rad/s to 131 rad/s in 0.43 seconds. What is the angular

e-lub [12.9K]

I think it's 39.53

(please do calculate it. I am not completely sure)

5 0

3 years ago

Read 2 more answers

What’s the difference between applied and pure research

Ket [755]

Answer:

Pure research is conducted without any specific goal in mind. Applied research is conducted in order to solve a specific and practical problem, unlike pure research. Hope this helped.

8 0

3 years ago

An object is 10 cm from the mirror, its height is 1 cm and the focal length is 5 cm. What is the image height? (Indicate the obj

boyakko [2]

The image height is -10 cm (the image is upside down)

Explanation:

We can solve the problem by using the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where:

f = 5 cm is the focal length of the mirror

p = 10 cm is the distance of the object from the mirror

q is the distance of the image from the mirror

Solving for q, we find:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\\\rightarrow q= 10 cm$

So, the distance of the image from the mirror is 10 cm.

Now we can find the image height by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the height of the image

y = 1 cm is the height of the object

and using

p = 10 cm

q = 10 cm

We find the size of the image:

$y' = -\frac{qy}{p}=-\frac{(10)(1)}{10}=-10 cm$

where the negative sign indicates that the image is upside down.

#LearnwithBrainly

8 0

3 years ago